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Topic: Help: a direct proof of Independence of AC (Read 851 times) |
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beibeibee
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Help: a direct proof of Independence of AC
« on: Nov 2nd, 2004, 1:49am » |
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In Kunen’s Set Theory, P245, there is a famous problem (settled by Cohen):
show that consis(ZF) \to consis(ZF+\neg AC).
I know a proof of using symmetric model, which however is involved in the Boolean-valued models.
Kunen gives a hint to use p.o. straightforwardly to construct a submodel of generic model, in which ZF holds but AC fails. I have tried to handle this proof in these days, but I find it is out of my control  . Is there anyone telling me that the main idea of such a proof. Many thanks!
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kiochi
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Re: Help: a direct proof of Independence of AC
« Reply #1 on: Oct 31st, 2006, 5:18pm » |
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You can use cohen forcing to construct a model in which AC fails. Take a model R so that:
M is a submodel of R is a submodel of M[G]
where M is a (countable) transitive model of ZFC and M[G] is a generic extension of M.
Many suitable models exist. For example, take M=L and extend M by adding countably many Cohen generic reals. Then it's fairly easy to construct a model R so that R is a transitive model of ZF in which the reals are not well-orderable.
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Michael Dagg
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Re: Help: a direct proof of Independence of AC
« Reply #2 on: Nov 1st, 2006, 5:37pm » |
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Example?
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Regards,
Michael Dagg
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kiochi
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Re: Help: a direct proof of Independence of AC
« Reply #3 on: Dec 24th, 2006, 4:59am » |
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Sorry, was gone for a while. I'll use a bit (but try to limit the use) of TeX markup, hope that's OK.
So we w.t.s con(ZF) ==> con(ZF+¬AC).
Assuming con(ZF) we can take c.t.m. M of ZFC (given a model of ZF, theory of constructibility implies there also exists model of ZFC).
Now define a poset B := {f:f is a finite fcn, f\subseteq \omega_1 \times 2} ordered by \supseteq (with largest element \emptyset).
Let G be B-generic over M.
define a fcn * by recursion, setting x*={(y*,1):y \in x}
Constructible Heirarchy (let D is the definable powerset operation, tc is the transitive closure):
L_{0}^{AB} = tc(A)
L_{a+1}^{AB}= D(L_{a}^{AB}, C)
L_{a}^{AB} = \bigcup_{b <a} L_{b}^{AB} for limit ordinals a
L_{a}(A) = L_{a}^{{B}\emptyset}
lL(A) = \bigcup_{a an ordinal} L_{a}(A)
It's straightforward (if tedious) to show that lL will be a model of ZF.
Now let N = lL(Pwrst(\omega))^{rel M[G]}
(that is, relativized to the model M[G])
So N is a model of ZF, and in fact AC fails in N.
Suppose to the contrary that AC holds in N. Let k = |Pwrst(\omgea)|.
(again the ^{rel __} means relativized to __).
Ok, so the statement (k*=|Pwrst(\omega)|)^{rel lL(Pwrst(\omega)} holds in M[G], so let p \in G be s.t. p forces (|k*|=|Pwrset(\omega)|)^{rel lL(Pwrst(\omega)}.
So 1 forces (|k*|=|Pwrst(\omega)|)^{rel lL(Pwrst(\omega)} with respect to our poset B.
OK, so this really takes some proof  , but I haven't let details get in my way so far, so let's not start 
Now define a new poset. Let j be the least cardinal greater than |k|, now set D := {f:f is a finite fcn, f\subseteq \j \times 2} (ordered like B)
Now 1 forces (|k*|=|Pwrst(\omega)|)^{rel lL(Pwrst(\omega)} with respect to D.
Let H be D-Generic over M. Then (|k*|=|Pwrst(\omega)|)^{rel lL(Pwrst(\omega)} holds in M[H]. So a bijection exists in M[H] from Pwrst(\omega) to k. But |Pwrst(\omega)| ? j, contra.
whew. Yea I left out a bunch of details let me know if you need any in particular ... but that's the rough sketch.
--k
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| « Last Edit: Dec 24th, 2006, 9:48am by kiochi » |
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